Binary Floats Represented as Decimal Numbers

Question

Not all decimal numbers can be represented exactly using binary floats.

http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html

There are two reasons why a real number might not be exactly representable as a floating-point number. The most common situation is illustrated by the decimal number 0.1. Although it has a finite decimal representation, in binary it has an infinite repeating representation.

What about the other way around? Can every single IEEE 754 float be represented exactly using a decimal number, if enough digits are used?

@harold Some double-precision numbers require about 750 decimal digits (more or less 52 + 1022 * (1 - log10(2))) — Pascal Cuoq
@harold I'm not counting them when I say 52 + 1022 * (1 - log10(2)), but if I wanted to count them it would be 52 + 1022. Not counting leading zeroes is easy: each binary digit in the significand and each unit of negative exponent add the need for one decimal digit. — Pascal Cuoq
@harold This person says 767. The “1 - log10(2)” thing is a rough approximation, not an exact science. stackoverflow.com/a/17245451/139746 — Pascal Cuoq

Pascal Cuoq Pascal Cuoq · Accepted Answer · 2015-10-04T17:25:45

Yes, every finite IEEE 754 float be represented exactly using a decimal number, if enough digits are used.

Each additional binary digit of precision requires at most one additional decimal digit of precision to represent exactly.

For instance:

0.1b   -> 0.5
0.01b  -> 0.25
0.11b  -> 0.75
0.001b -> 0.125

A double-precision (binary64) number between 1 and 2 requires only 52 decimal digits after the dot to be represented exactly:

#include <stdio.h>

int main(void) {
  printf("%.55f\n", 1.1);
}

Result:

1.1000000000000000888178419700125232338905334472656250000

It's all zeroes after the four displayed at the end of the representation above. 1.100000000000000088817841970012523233890533447265625 is the exact value of the double nearest to 11/10.

As pointed out in the comments below, each additional unit of magnitude for a negative exponent also requires one additional decimal digit to represent exactly. But negative exponents of a large magnitude have leading zeroes in their decimal representations. The smallest subnormal number would have 1022 + 52 decimal digits after the dot, but the first nearly 1022*log₁₀(2) of these digits would be zeroes.

Binary Floats Represented as Decimal Numbers

1 Answers